2. +
LEAST MASTERED SKILLS
Solving Quadratic Equation
Sub Tasks
Identifying quadratic equations
Rewriting quadratic equations to its
standard form
Factor trinomials in the form x2 + bx + c
Determine roots of quadratic equation
ax2 + bx + c = 0, by factoring
3. +
Overview
A quadratic equation in one variable is a
mathematical sentence of degree 2 that can
be written in the following form
ax2 + bx + c = 0,
where a, b, and c are real numbers and
a ≠ 0.
How are quadratic equations used
in solving real – life problems and in
making decisions?
Many formulas used in
the physical world are
quadratic in nature since they
become second-degree
equations when solving for one
of the variables. Likewise,
many word problems require
the use of the quadratic
equation.
At the enrichment card,
we will consider one of the
common use of the quadratic
equations.
4. + Activity Card # 1
__________ 1. 3m + 8 = 15
__________ 2. x2 – 5x – 10 = 0
__________ 3. 2t2 – 7t = 12
__________ 4. 12 – 4x = 0
__________ 5. 25 – r2 = 4r
Quadratic or Not Quadratic?
Direction. Identify which of the following equations
are quadratic and which are not.
Write QE if the equations are quadratic and NQE if
not quadratic equation.
5. +
Activity Card # 2
Set Me to Your Standard!
Direction. Write each quadratic equation in standard
form, ax2 + bx + c = 0.
1. 3x – 2x2 = 7 ____________________
2. 5 – 2r2 = 6r ____________________
3. 2x(x – 3) = 15 ____________________
4. (x + 3)(x + 4)= 0 ____________________
5. (x + 4)2 + 8 = 0 ____________________
6. + Activity Card # 3
What Made Me?
We learned how to multiply two binomials as follows:
factors
(x+2)(x+6) = x2 + 6x + 2x + 12 = x2 + 8x + 12.
terms
M u l t i p l y i n g
factorsterms
F a c t o r i n g
x2 + 8x + 12 = (x + 2)(x + 6)
In factoring, we reverse the operation
The following will enable us to see how a trinomial factors.
x2 + 8x + 12 = (x + 2)(x + 6)
12 = 2 (6)
8 = 2 + 6
Product
Sum
Study Tip
Alternate
MethodYou can use the opposite
of FOIL to factor
trinomials. For instance,
consider
Example 1.1
x2+ x – 12
(x + )(x + )
Try factor pair of -12 until
the sum of the products of
the Inner and Outer terms
is x.
7. +
In general, the trinomial x2 + bx + c will factor only if there are two
integers, which will we call m and n, such that m + n = b and
m(n) = c.
Sum Product
m + n m(n)
x2 + bx + c = (x + m)(x + n)
1. a2 + 11a + 18 m + n = 11 m(n) = 18
2 + 9 = 11 2(9) = 18
The m and n values are 2 and 9. the factorization is,
a2 + 11a + 18 = (x + 2) (x + 9)
2. b2 – 2b – 15 m + n = - 2 m(n) = - 15
3 + (-5) = - 2 3(-5) = - 15
The m and n values are 3 and - 5. the factorization is,
b2 – 2b – 15 = (x + 3) (x – 5)
8. +
Factor the following trinomial in the form x2 + bx + c.
x2 + bx + c m + n m(n) (x + m)(x + n)
x2 + 4x – 12 6 + (-2) 6(-2) (x + 6)(x – 2)
w2 – 8w + 12
x2 + 5x - 24
c2 + 6c + 5
r2 + 5r – 14
x2 + 9x + 20
After learning how to factor trinomial in the form x2 + bx + c,
we will now determine roots of a quadratic equation using factoring.
9. +
Activity Card # 4 Factor then Solve!
Some quadratic equations can be solved easily by factoring. To solve each equations, the
following procedures can be followed.
1. Transform the quadratic equation into standard form if necessary.
2. Factor the quadratic expression.
3. Set each factor of the quadratic expression equal to 0.
4. Solve each resulting equation.
Example. Find the solution of x2 + 9x = -8 by factoring.
a. Transform the equation into standard form
x2 + 9x = -8 x2 + 9x + 8 = 0
b. Factor the quadratic expression
x2 + 9x + 8 = 0 (x + 1)(x +8) = 0
c. Set each factor equal to 0.
(x + 1)(x + 8) = 0 x + 1 = 0 ; x + 8 = 0
d. Solve each resulting equation.
x + 1 = 0 x + 1 – 1 = 0 -
1
x = - 1
x + 8 = 0 x + 8 – 8 = 0 - 8
x = - 8
11. +
Choose the letter that best answer the question.
__________ 1. A polynomial equation of degree 2 that can be written in the form
𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0, where a, b and c are real numbers, and a≠ 0.
a. Linear Equation
b. Linear Inequality
c. Quadratic Equation
d. Quadratic Inequality
__________ 2. Which of the following is a quadratic equation?
a. 𝑥2
+ 2𝑥 + 1 = 0 c. 4𝑏 − 2 = 12
b. 𝑦3
− 1 = 0 d. 𝑥 = 5
__________ 3. The following are quadratic equation written in standard form except
a. 3𝑡 − 7 = 2 c. 2𝑟2
+ 4𝑟 − 1
b. 𝑠2
+ 5𝑠 − 4 = 0 d. 𝑥2
+ 2𝑥 = 2
__________ 4. What is the standard form of the quadratic equation 𝑥2
+ 4𝑥 = 4?
a. 𝑥2
+ 4𝑥 + 4 = 0 c. 𝑥2
− 4𝑥 + 4 = 0
b. 𝑥2
− 4𝑥 − 4 = 0 d. 𝑥2
+ 4𝑥 − 4 = 0
12. +
__________ 5. What are the factors of the trinomial 𝑠2
+ 8𝑠 + 15?
a. (𝑠 – 3)(𝑠 – 5) c. (𝑠 + 3)(𝑠 + 5)
b. (𝑠 + 3)(𝑠 – 5) d. (𝑠 – 3)(𝑠 + 5)
__________ 6. If one of the factor of the trinomial 𝑥2
+ 10𝑥 + 25 is 𝑥 + 5, what is the
other factor?
a. (𝑥 – 5) c. (𝑥 – 2)
b. (𝑥 + 2) d. (𝑥 + 5)
__________ 7. What is the roots of the quadratic equation 𝑥2
+ 9𝑥 + 8 = 0?
a. 𝑥 = −1; 𝑥 = − 8 c. 𝑥 = − 1; 𝑥 = 8
b. 𝑥 = 1; 𝑥 = − 8 d. 𝑥 = 1; 𝑥 = 8
__________ 8. The roots of the quadratic equation are – 5 and 3. Which of the
following quadratic equations has these roots?
a. 𝑥2
− 8𝑥 + 15 = 0 c. 𝑥2
− 2𝑥 − 15 = 0
b. 𝑥2
+ 8𝑥 + 15 = 0 d. 𝑥2
+ 2𝑥 − 15 = 0
__________ 9. Which of the following term must not be equal to 0 in a quadratic
equation?
a. 𝑎𝑥2
b. 𝑏𝑥 c. 𝑐 d. 0
__________ 10. In the quadratic equation 𝑥2
− 4𝑥 + 3 = 0, one of the roots is 1. What
is the other root?
a. 3 b. −1 c. − 3 d.
1
3
13. +
*The length of a rectangle is 5 cm more than its width and the area is 50 square cm. Find the
length and the width of the rectangle.
Solution: w
w + 5
w + 5
Use the formula 𝐴𝑟𝑒𝑎 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑡𝑖𝑚𝑒𝑠 𝑤𝑖𝑑𝑡ℎ 𝑜𝑟 𝐴 = 𝑙𝑤 and the fact that the area is 50 square
cm to set up an algebraic equation.
𝐴𝑟𝑒𝑎 = 𝑙𝑒𝑛𝑔𝑡ℎ (𝑤𝑖𝑑𝑡ℎ)
50 = 𝑤 + 5 (𝑤)
Simplifying it, we notice that the equation is a quadratic equation.
50 = 𝑤2
+ 5𝑤
By using the concepts of solving quadratic equation by factoring, we get
𝑤2
+ 5𝑤 − 50 = 0
(w + 10) (w – 5) = 0
w + 10 = 0 w – 5 = 0
w = - 10 w = 5
At this point, we have two possibilities for the width of the rectangle, However, since w = - 10 is
impossible to be a width, choose the positive solution, w = 5. Back substitute to find the length,
length, w + 5 = 5 + 5 = 10.
Answer: The width is 5 feet and the length is 10 feet.
(Note: It is important to include the correct unit in the presentation of the answer. Make sure to indicate that the
width is 5 feet and the length is 10 feet.)
5 cm more than the width
14. +
Now it’s your turn…!
Problem:
The floor of a rectangular room has a length that is 4 feet more than twice its
width. If the total area of the floor is 240 square feet, then find the dimensions of
the floor. (Note: The dimensions of the floor is the length and width of the floor.)
Answer:
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15. +
Learner’s Material – Mathematics IX, First Edition pp. 27 – 34
Holiday, Berchie. et. al. ALGEBRA 2. USA. The McGraw – Hill
Companies, 2008. pp. 253 – 256
Wesner, et. al. ELEMENTARY ALGEBRA with APPLICATIONS.
Bernard J. Klein Publishing, 2006 pp. 152 – 156
16. +
Activity Card # 1 Quadratic or Not Quadratic?
1. NQE
2. QE
3. QE
4. NQE
5. QE
Activity Card # 2 Set Me to Your Standard
1. - 2x2 + 3x – 7 = 0 or 2x2 – 3x + 7 = 0
2. - 2r2 – 6r + 5 = 0 or 2r2 + 6r – 5 = 0
3. 2x2 – 6x – 15 = 0
4. x2 + 7x + 12 = 0
5. x2 + 8x + 24 = 0
Activity Card # 3 What Made Me?
x2 + bx + c m + n m(n) (x + m) (x + n)
w2 – 8w + 12 - 6 + 2 -6(2) (w – 6)(w + 2)
x2 + 5x – 24 8 + (-3) 8(-3) (x + 8)(x – 3)
c2 + 6c + 5 5 + 1 5(1) (c + 5)(c + 1)
r2 + 5r – 14 7 + (-2) 7(-2) (r + 7)(r – 2)
x2 + 9x + 20 5 + 4 5(4) (x + 5)(x + 4)
Activity Card # 4
1. x2 + 8x + 16 = 0 (x + 4)(x + 4) = 0
x + 4 = 0
x + 4 – 4 = 0 – 4
x = - 4
2. x2 – 5x – 14 = 0 (x – 7)(x – 2) = 0
x – 7 = 0 x – 2 = 0
x – 7 + 7 = 0 + 7 x – 2 + 2 = 0 + 2
x = 7 x = 2
3. y2 + 9y + 20 = 0 (y + 5)(y + 4) = 0
y + 5 = 0 y + 4 = 0
y + 5 – 5 = 0 – 5 y + 4 – 4 = 0 – 4
y = - 5 y = - 4
4. b2 – 10b + 21 = 0 (b – 7)(b – 3) = 0
b – 7 = 0 b – 3 = 0
b – 7 + 7 = 0 + 7 b – 3 + 3 = 0 + 3
b = 7 b = 3
Assessment Card
1. c 6. d
2. a 7. a
3. a 8. d
4. d 9. a
5. c 10. a
Enrichment Card
Answer:
The width is 10
feet and the length
is 24 feet.